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This is the second of our many quizzes on Magnetic force. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Magnetic force Quiz 2

physics

Suppose a proton travels with speed of light $(3\times {{10}^{8}}\,m/s)$ in a circular orbit of x-y plane under the effect of magnetic field of strength 1 tesla along the positive z direction. When electric field of $1\,V{{m}^{-1}}$ is applied along the positive $y$ direction, the center of the circular orbit- 

Direction of $\overset{\to }{\mathop{B}}\,$,$\overset{\to }{\mathop{E}}\,$ and the particle velocity will be perpendicular to each other. Since it is given that the magnetic field is in $z$ direction and electric field is in $y$ direction, we can easily find the direction of the velocity of the charge. 

  $\text{The drift velocity is given by}\\   \overset{\to }{\mathop{{{V}_{E}}}}\,=\dfrac{\overset{\to }{\mathop{{{E}_{\bot }}}}\,\times \overset{\to }{\mathop{B}}\,}{{{B}^{2}}}\\  \Rightarrow\overset{\to }{\mathop{{{V}_{E}}}}\,=\dfrac{\hat{y}\times \hat{z}}{1}\\  \therefore\overset{\to }{\mathop{{{V}_{E}}}}\,=1\text{ }\left( {\hat{i}} \right)m/s\\  \text{Therefore, the particle will move at}\,1m{{s}^{-1}}\,\text{along the positive x direction}$  

Which of the following phenomena is based on the effects of Magnetic Field?

Aurora lights are also known as northern lights and they are mostly visible near the poles. We must know that the magnetic field at poles is higher at the poles than at the equator.  

  As solar winds from our sun hit Earth, the charged particles in the solar wind interact with the Earth's magnetic field, releasing a massive amount of energy that we see as bright lights known as Aurora lights.

 A charge Capacitor “C” is connected in series with an inductor “L”. When the displacement current reduces to zero, the energy of the $LC$ Circuit is

When the capacitor becomes fully charged in an $LC$ circuit, the energy stored in an inductor is stored in the form of magnetic field. The displacement current density is given by the formula $J={{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial E}{\partial t}$. 

 $\text{The displacement current density is given by}\\   J={{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial E}{\partial t}\\  \text{Let “A” be the area of plate of capacitor, now displacement current is}\\ {{I}_{D}}=JA={{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial E}{\partial t}\\  \Rightarrow{{I}_{D}}=0={{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial E}{\partial t}=0\\  \text{Therefore,}\,\dfrac{\partial E}{\partial t}=0\\  E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\\  \Rightarrow E =\dfrac{Q}{2{{\varepsilon }_{0}}A}\\ \Rightarrow\dfrac{\partial E}{\partial t}=\dfrac{1}{2{{\varepsilon }_{0}}A}\dfrac{\partial Q}{\partial t}\\ \Rightarrow 0=\dfrac{1}{2{{\varepsilon }_{0}}A}\times I\\  \therefore I=0\\  \text{Since, the capacitor is fully discharged so the total energy is stored in an inductor in the form of magnetic field.}$

The magnitude of vector potential inside the cylinder at a distance $R$ from its axis is 

We can find the magnitude of vector potential inside the cylinder by using Biot-savart's law which states the magnetic intensity at any point due to a steady current in an infinitely long straight wire is directly proportional to the current and inversely proportional to the distance from point to wire. 

 $\text{We know that,}\\  \oint{{\vec{A}}}.d\vec{l}=\int{\vec{B}.d\vec{s}}\\  \Rightarrow A.2\pi r={{\mu }_{0}}\sigma \omega R.\pi {{r}^{2}}\\  \therefore A=\dfrac{1}{2}{{\mu }_{0}}\sigma R\omega r$

 Find the work done by a particle which is travelling with a constant speed $(\text{say v})$ in a magnetic field $\overset{\to }{\mathop{B}}\,$ 

When a particle moves with a constant speed in a magnetic field, the force acting on the particle is only due to the magnetic field. This force on a charged particle placed in a magnetic field is given by the product of that charge to the cross product of velocity of that particle and the magnetic field. 

 $  \text{The force acting on a particle due to magnetic field an be given by}\\  \overset{\to }{\mathop{F}}\,=q(\overset{\to }{\mathop{V}}\,\times \overset{\to }{\mathop{B}}\,)\\  \text{Now, we know that work done is defined as the product of Force with displacement,}\\  W=\int{\overset{\to }{\mathop{F}}\,}.\overset{\to }{\mathop{dl}}\\  \Rightarrow W=q\int{(\overset{\to }{\mathop{V}}\,\times \overset{\to }{\mathop{B}}\,)}.\overset{\to }{\mathop{dl}}\\  \text{Since,}\,\overset{\to }{\mathop{V}}\,\text{and}\,\overset{\to }{\mathop{dl}}\,\text{are parallel and}\,\overset{\to }{\mathop{V}}\,\times \overset{\to }{\mathop{B}}\,\text{is perpendicular to}\,\overset{\to }{\mathop{dl}}\\  \therefore(\overset{\to }{\mathop{V}}\,\times \overset{\to }{\mathop{B}}\,).\overset{\to }{\mathop{dl}}\,=0\\  \text{Hence, work done will also be zero.}$